Solve for $n$, $ -\dfrac{n - 2}{n + 3} = \dfrac{1}{3n + 9} - \dfrac{3}{n + 3} $
Solution: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n + 3$ $3n + 9$ and $n + 3$ The common denominator is $3n + 9$ To get $3n + 9$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{n - 2}{n + 3} \times \dfrac{3}{3} = -\dfrac{3n - 6}{3n + 9} $ The denominator of the second term is already $3n + 9$ , so we don't need to change it. To get $3n + 9$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{3}{n + 3} \times \dfrac{3}{3} = -\dfrac{9}{3n + 9} $ This give us: $ -\dfrac{3n - 6}{3n + 9} = \dfrac{1}{3n + 9} - \dfrac{9}{3n + 9} $ If we multiply both sides of the equation by $3n + 9$ , we get: $ -3n + 6 = 1 - 9$ $ -3n + 6 = -8$ $ -3n = -14 $ $ n = \dfrac{14}{3}$